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convergance of sum of Ln(n)/n
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the sum diverges actually
so we would compare it to 1/n
yes, ln(n)/n > 1/n, since 1/n diverges, by comparison test, ln(n)/n also diverges
ok so 1/n diverges by the harmonic series test
yea
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can u explain tht pls
The comparison test states that, if an > bn for all natural n and sum (bn) diverges, then sum (an) diverges
is this the same thing for (ln(n))^2/n
cause (ln(n))^2 > 1^2
yes
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kk thank u!
one minor point: your inequality is not true for all natural numbers...so you should point out that \[\ln(n)>1 \text{ for }n\ge3\] which is all you need. you don't really care what happens at the beginning of the sequence, only what happens after a finite sequence of terms.
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